From: Alfred Nykolyn Subject: Re: A Geometry problem Date: Thu, 22 Nov 2001 13:47:03 -0700 Newsgroups: sci.math Summary: Adventitious angles problems (solving triangles with "unexpectedly" integral solutions) Here is a little twist to the Adventious Angles problem which appeared in Kvant some time ago: [Correction] Given an isosceles triangle ABC in which AB is equal to AC and the angle BAC equals 20 degrees. (so BC is the base of the triangle) A point D is defined on AC so that AD is equal to BC. What is the size of angle DBC? ============================================================================== From: Lewis Mammel Subject: Re: A Geometry problem Date: Fri, 23 Nov 2001 07:02:21 GMT Newsgroups: k12.ed.math,sci.math,uk.education.maths Dave Wilson wrote: > > In article , > "Zarang" wrote: > > >Many thanks to all who replied to my posting. I have now completed the > >problem. > >I had not realised that it had quite a history and that under the title of > >"Adventitious angles " there were considerable web references to the > >problem. I am much obliged to all who either replied to the group or > >e-mailed me the solution. > > Do, please, share some of the details of approaches, and, perhaps, some of > the useful URLs (I know, now, that I can chuck AA into a search engine, but > your thoughts on useful ones will assist) The first ( top-most ) google match on [ adventitious angles ] : http://www.dcs.st-and.ac.uk/~ad/mathrecs/advent/advent.html gives a link to a geometric proof, and mentions that "a number of trig. solutions are possible." I'll mention the one that occurred to me, as it's fairly simple: By applying the law of sines to the triangle APQ we get AP/AQ = sin x / sin y ; x = angle AQP, y = angle APQ by applying the law of sines to triangle APC we get AP/AC = sin 30 / sin 130 and correspondingly with triangle AQC we get AQ/AB = sin 20 / sin 140 then angle BQP = 30 iff x = 110, y = 50 iff sin 110 / sin 50 = sin 30 * sin 140 / sin 20 / sin 130 iff sin 70 / sin 50 = sin 30 * sin 40 / sin 20 / sin 50 iff cos 20 * sin 20 / sin 30 = sin 40 but since sin 30 = 1/2, this is true by the double angle formula. Lew Mammel, Jr. -- submissions: post to k12.ed.math or e-mail to k12math@sd28.bc.ca private e-mail to the k12.ed.math moderator: kem-moderator@thinkspot.net newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ============================================================================== From: "Zarang" Subject: Re: A Geometry problem Date: Fri, 23 Nov 2001 16:54:17 GMT Newsgroups: k12.ed.math,sci.math,uk.education.maths This posting seems to have aroused considerable interest and I am grateful to all of those who either responded to the groups or to me by e-mail. I should especially like to thank Dave Rusin and R.S Tiberio for their useful references or advice. It seems that this problem was originally proposed by F.M. Langley and belongs to a set of problems known as "Adventitious Angles". A Google search found 1,110 references to Adventitious Angles. However the most useful sites which contain solutions or valuable clues are:- http://cmgm.stanford.edu/~ahmad/solaug96.html#solution4 http://cmgm.stanford.edu/~ahmad/puzaug96.html http://www.dcs.st-and.ac.uk/~ad/mathrecs/advent/soln1.html I hope this information will assist those who are searching for solutions. Zarang. -- submissions: post to k12.ed.math or e-mail to k12math@sd28.bc.ca private e-mail to the k12.ed.math moderator: kem-moderator@thinkspot.net newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html